Append to a pipe and pass on?

Question

I have a simple bash function dividing two numbers:

echo "750/12.5" | bc

I'd like to take the output from bc and append /24 and pipe said result to another instance of bc.

Something like:

echo "750/12.5" | bc | echo $1 + "/24" | bc

Where $1 is the piped result.

P.S. I realize I could just do echo "750/12.5/24" | bc my question is more in regards to the appending of text to a pipe result.

Answer 1

I achieved this using sed and replacing the end of the line:

echo "my text" | sed 's/$/ more text/'

Returns:

my text more text

Your example:

echo "750/12.5" | bc | sed 's/$/\/24/' | bc

Answer 2

In the simplest of the options, this does append to the pipe stream:

$ echo "750/12.5" | { bc; echo "/24"; }
60
/24

However that has an unexpected newline, to avoid that you need to either use tr:

$ echo "750/12.5" | { bc | tr -d '\n' ; echo "/24"; }
60/24

Or, given the fact that a command expansion removes trailing newlines:

$ printf '%s' $( echo "750/12.5" | bc ); echo "/24"
60/24

But probably, the correct way should be similar to:

$ echo "$(echo "750/12.5" | bc )/24"
60/24

Which, to be used in bc, could be written as this:

$ bc <<<"$(bc <<<"750/12.5")/24"
2

Which, to get a reasonable floating number precision should be something like:

$ bc <<<"scale=10;$(bc <<<"scale=5;750/12.5")/24"
2.5000000000

Note the need of two scale, as there are two instances of bc.

Of course, one instance of bc needs only one scale:

$ bc <<<"scale=5;750/12.5/24"

In fact, what you should be thinking about is in terms of an string:

$ a=$(echo "750/12.5")        # capture first string.
$ echo "$a/24" | bc           # extend the string
2

The comment about scale from above is still valid here.

Answer 3

Alternative approach:

TL;DR: Use cat to concatenate with stdin (-) as an argument. Note that it'll newline-separate your concatenated streams; remove if undesired.

Unfortunately this approach is a bit cumbersome for your particular use case since bc fails if there's a newline in the middle of that expression:

$ echo "750/12.5" | bc | cat - <(echo "/24") | bc
60
(standard_in) 2: parse error

You can work around that by deleting the newline from the original bc result:

$ echo "750/12.5" | bc | tr -d '\n' | cat - <(echo "/24") | bc
2

However, I came upon this question by googling "bash append to pipe", and in my case I actually do want that newline between the concatenations. So for those here for the same reason, use | cat - <(...):

$ echo foo | cat - <(echo bar)
foo
bar
$ echo foo | cat - <(echo bar) | sort | cat - <(echo last_line)
bar
foo
last_line
$ echo foo | cat - <(echo bar) | sort | cat <(echo header) - <(echo last_line)
header
bar
foo
last_line

Answer 4

Something like this seems to work using xargs:

$ echo "750/12.5" | bc |xargs -I{} echo "{}+20" |bc
80

Or even:

$ echo "750/12.5" | bc |echo "$(</dev/stdin)+20" |bc
80

And why not :

$ bc <<<20+$(bc <<<"750/12.5")
80